Si $f(0)=0$$f(1)=1$, demuestran que, a $\int_0^1 \left |f'(x)-f(x) \right |dx\geq e^{-1}$

Question

Deje $f$ ser una función derivable en a $[0,1]$ tal que $f(0)=0$$f(1)=1$. Si $f'$ es continua, demostrar que $$\int_0^1 \left |f'(x)-f(x) \right |dx\geq e^{-1}$$

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El progreso

Dejo $h(x)=e^{-x}f(x)$, por lo que $$\int_0^1 \left |f'(x)-f(x) \right |dx=\int_0^1e^x\left |h'(x) \right |dx$$ But I can't continue from here.. Maybe $$\int_0^1e^x\left |h'(x) \right |dx \geq \max_{x\in [0,1]}e^x\left |h'(x) \right |$$ ayuda?

Answer 1

Tenemos, $$\begin{align}\int_0^1|f'(x)-f(x)|dx &= \int_0^1|f'(x)e^{-x}-f(x)e^{-x}|e^xdx \\&\geq \int_0^1\left(f'(x)e^{-x}-f(x)e^{-x}\right) dx\\&= \int_0^1 \frac{d\left(f(x)e^{-x}\right)}{dx}dx\\&= f(1)e^{-1}-f(0)e^{0}\\&=\frac{1}{e}\end{align}$$